In linear algebra, a vector space $V$ over a field $\mathbb{F}$ has a dual vector space $V^*$, and it's said somewhere that $V^*$ comes "for free". By "for free", I guess it means that there is nothing else to be required other then $V$ & $ \mathbb{F}$ for obtaining $V^*$.
In the similar setting, it seems that given two vector spaces $V, W$ over $\mathbb{F}$, the vector space $\hom(V,W)$ also kind of "comes for free".
I am curious on what separates $V^*$ (i.e., $\hom(V,\mathbb{F})$) apart from $\hom(V,W)$ to give it a dedicated name "dual space" (besides $\dim V=\dim V^*$)?
Or, as the title shows, is $\hom(V,W)$ a dual space of something else? If not, why?
First a brief remark about your first paragraph: The data specifying the vector space structure of a vector space $V$ includes the underlying field, so in that sense $V^*$ can be defined just using the vector space $V$.
Now, expanding Douglas' remark in the comments, for f.d. vector spaces $V, W$ over a given field $\Bbb F$ we have $$\boxed{\operatorname{Hom}(V, W)^* \cong (W \otimes V^*)^* \cong W^* \otimes V^{**} \cong W^* \otimes V \cong V \otimes W^* \cong \operatorname{Hom}(W, V)} .$$
We can describe this isomorphism explicitly: Define the linear map $$\Phi : \operatorname{Hom}(W, V) \to \operatorname{Hom}(V, W)^*$$ by $$\Phi(T)(S) := \operatorname{tr}(S \circ T) .$$ By definition if $T \in \ker \Phi$, then $\operatorname{tr}(S \circ T) = 0$ for all $S \in \operatorname{Hom}(V, W)$, so $T = 0$, and thus $\Phi$ is an isomorphism.
In the case $W = \Bbb F$, $S \in \operatorname{Hom}(V, \Bbb F) = V^*$ and $T \in \operatorname{Hom}(\Bbb F, V)$, but we can naturally identify this latter space with $V$ via the map $f \mapsto f(1)$, so $\Phi(S)(T)$ is a map $\Bbb F \to \Bbb F$, which we can identify with its trace, and hence under these canonical identifications $\Phi$ specializes to the pairing $V^* \times V \to \Bbb F$, $(\alpha, X) \mapsto \alpha(X)$, defining the dual $V^*$ of a vector space $V$.